Derivative f’ of function f(x)=arctan x is: f’(x) = 1 / (1 + x²) for all x real. To show this result, we use derivative of the inverse function tan x.

Derivative of arctan x

Derivative $f’$ of function $f(x)=\arctan{x}$ is: \(\forall x \in \mathbb{R} ,\quad f'(x) = \dfrac{1}{1+x^2}\)


Remember that function $\arctan$ is the inverse function of $\tan$ :

\[\left(f^{-1} \circ f\right)=\left(\tan \circ \arctan\right)(x)=\tan(\arctan(x))=x\]

Using result of derivative of inverse functions+ we have:


Taking : $g^{-1}=f=\arctan$ and $g=f^{-1}=\tan$, we have: \(f^{\prime}(x)=\frac{1}{\tan^{\prime}(f(x))}\)

We have:

Derivative of tangent function $g(x)=\tan x$ is: \(\forall x \neq \dfrac{\pi}{2}+k\pi, k \in \mathbb{Z}, \quad g'(x) = 1+\tan ^{2} x\)

So, we have:

\[\begin{aligned} f^{\prime}(x)&=\frac{1}{\tan^{\prime}(f(x))}\\ &=\frac{1}{1+\tan^2(f(x))}\\ &=\frac{1}{1+\tan^2(\arctan x)}\\ \end{aligned}\]

Now by defiinition: \(\left(f^{-1} \circ f\right)=\left(\tan \circ \arctan\right)(x)=\tan(\arctan(x))=x\)

We conclude that: \(\forall x \in \mathbb{R} ,\quad f'(x) = \dfrac{1}{1+x^2}\)